2.0 g of a bleaching powder sample is suspended in 100 mL of water. A 10 mL aliquot of this suspensi — Redox Reactions and Volumetric Analysis Chemistry Question
Question
2.0 g of a bleaching powder sample is suspended in 100 mL of water. A 10 mL aliquot of this suspension is treated with excess $KI$ and acidic medium. The liberated iodine requires exactly 10 mL of 0.2 N $Na_2S_2O_3$ to reach the end point. What is the percentage of available chlorine in the bleaching powder?
💡 Solution & Explanation
For the 10 mL aliquot, milli-equivalents of $Cl_2$ = meq of $I_2$ = meq of Hypo = $10 \times 0.2 = 2$ meq. Since the total suspension is 100 mL, the total milli-equivalents of $Cl_2$ in the 2.0 g sample = $2 \times (100 / 10) = 20$ meq = 0.02 equivalents. The equivalent weight of $Cl_2$ is 35.5 (MW = 71, n-factor = 2). Mass of available $Cl_2$ = $0.02 \times 35.5 = 0.71$ g. Percentage of available $Cl_2$ = $(0.71 / 2.0) \times 100 = 35.5\%$.